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Help with intergration

  1. Apr 14, 2005 #1

    liz

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    URGENT help with intergration

    im doing some coursework where im investigating ways of finding the area under a graph and i dont really understand intergration. please could someone explain it really thoroughly, id really appreciate it. thank you in advance to anyone who can help.
     
  2. jcsd
  3. Apr 14, 2005 #2
    Well, integration (between two integration-boundaries a and b where a<b) really calculates the area under a given curve f(x). The start and end-point are the boundaries.

    [tex]\int_{a}^{b} f(x)dx[/tex] really means that you sum up (denoted by the integration sign, which is the continuous variant of the sigma-sommation sign) the surface of little rectangles. The f(x) denotes the height and the differential dx denotes the width. So f(x)*dx really is the surface, you see?

    This is in easy language the signification of an integral in one dimension. Mathematically all of this is formalised using Darboux-sums for the Riemann-integral. You can also use measure-theory to introduce the concept of integrals, which is then called the Lebesgue integral

    You know, that integration is the opposite of derivation
    http://en.wikipedia.org/wiki/Riemann_integral
    regards
    marlon
     
    Last edited: Apr 14, 2005
  4. Apr 14, 2005 #3
    An integral is a Riemann sum with the limit taken to infinite, (or dx-> 0). What it does is it makes little rectangles of width Delta X and height f(x) throughout an interval a to b. The sum of these rectangles approximates the area underneath, and as the number of rectangles goes to infinite, the sum becomes exact.

    Taking the limit of the riemann sum turns it into an integral ( I believe ).
     
  5. Apr 14, 2005 #4
    Conceptually, the integral really denotes the summation.

    It's the theory of Darboux sums and their minor and major rectangles of which the difference in area must evolve towards zero, that formally describe the theory begind integrals

    regards
    marlon
     
  6. Apr 14, 2005 #5
    Could you not say that [tex]\int_{a}^{b} y dx \approx \ \frac{1}{2} h [(y_0 \ + \ y_n) \ + \ 2(y_1 \ + \ y_2 \ + \ ... \ y_{n-1})] \ ,when \ h = \frac{b - a}{n}[/tex] ?

    This will show you can approximate integration even if you don't know what it is.

    The Bob (2004 ©)
     
  7. Apr 14, 2005 #6
    No, because how do you think this formula was constructed in the first place ?

    marlon
     
  8. Apr 14, 2005 #7

    dextercioby

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    And one more thing for liz

    INTEGRATION (from integer) (integral) and not intergration (as in inter grating)...:wink:

    At least the name get it right...

    Daniel.
     
  9. Apr 14, 2005 #8
    So critical. :)
     
  10. Apr 14, 2005 #9

    Imo

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    I suggest going to www.mathworld.com and typing in some of the keywords used in the replies. Specifically look up "Riemann Sum", "Integration" (at least on of these links will give you a brief history), and click on some of the links at the bottom of the page that catch your fancies.
     
  11. Apr 16, 2005 #10

    liz

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    thank you everyone for your help. and yes i know that it's spelt integration, i just put an extra r in automatically.


    isnt that formula that Bob sugested the trapezium rule?
    anyway thank you i am very grateful for the help form everyone.
     
  12. Apr 16, 2005 #11
    Yes, there are many ways to estimate an area.
     
  13. Apr 16, 2005 #12
    If you have any programming experience this might help. I used to think of it as a loop. Where the loop has a starting condition of A, and a stopping condition at B. Thus when you integrate a function from A to B, you are actually in a loop from A to B and the body of the loop (what its doing) is simply adding up every point the function "hits". The loop doesn't increment with integers though, instead it increments with infintesimals (really really tiny lengths). Thinking of it "looping with infintesimals" also will help you to remember the [tex] dx [/tex] in the integration notation with a problem like [tex] \int_{A}^{B} x^2 \,dx [/tex].


    I hope that doens't confuse you more. It is obviously not a thourough explanation, but you can find that, at many places on the internet. mathworld is an excellent place.
     
    Last edited: Apr 17, 2005
  14. Apr 17, 2005 #13

    liz

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    hey ive just been reading a pure maths textbook and it says at one point (i hope your familiar with integration stuff so i dont need to put all of it): from the definition, delta A / delta x tends to the derivative dA / dx. i know that delta means "an increase in ...", and d is used to mean "a change in ..." so what does that mean?
     
  15. Apr 17, 2005 #14
    Delta A / Delta X tends to dA/dX as Delta X goes to 0.

    delta X = x_2-x_1, the difference between two points, so if you think of a curve and a secant line drawn between any two points, as the secant line connects two points that are closer and closer together the line approaches the tangent line.
     
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